# Inverse Function

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• May 22nd 2011, 06:25 AM
adam_leeds
Self Inverse
(Headbang)

$f(x) = \frac{2}{x } - x$ so i have done $f(f(x)) -x = 0$ so $\frac{2}{\frac{2}{ x} -x} - (\frac{2}{ x} -x) - x = 0$ How does that equal $\frac{4{x}^{ 2} -4}{x(2-{x}^{ 2} ) }$ ???
• May 22nd 2011, 06:30 AM
SpringFan25
Shouldn't it be $f(f^{-1}(x)) - x =0$?

it may be easier to write
$y=\frac{2}{x} - x$
$yx=2 - x^2$

Which is a quadratic, so
$x=\frac{-y \pm \sqrt{y^2 + 8}}{2}$

and hence the inverse is
$f^{-1}(x)=\frac{-x \pm \sqrt{x^2 + 8}}{2}$
• May 22nd 2011, 06:33 AM
adam_leeds
Quote:

Originally Posted by SpringFan25
Shouldn't it be $f(f^{-1}(x)) - x =0$?

who knows what the name is called all i know is i have to do the formula i've put up on there.
• May 22nd 2011, 06:49 AM
SpringFan25
(not sure if you saw my previous post before i edited it to include the answer)

using your formula you should get the same answer as i posted:

$f(f^{-1}(x)) -x =0$
$\frac{2}{f^{-1}(x)} - f^{-1}(x) -x =0$

multiple everything by f^-1(x):
$2 - \left[f^{-1}(x)\right]^2 -xf^{-1}(x) =0$

this is a quadratic in $f^{-1}(x)$. the coefficients (a,b,c) are (-1,-x,2)

Solve with quadratic formula

$f^{-1}(x) = \frac{x \pm \sqrt{x^2 + 8}}{-2} = \frac{-x \pm \sqrt{x^2 + 8}}{2}$